In basketball, players often have to take free throws - Leaving Cert Mathematics - Question 8 - 2015

Here, we calculate the probability of being unsuccessful with the first two throws (which is 0.3 for each) and successful on the third throw (which is 0.8 when he is successful). Thus, we have:

$P(U, U, S) = 0.3 imes 0.3 imes 0.8 = 0.072$

The potential successful outcomes with the third throw can happen in the following ways:

1. S, S, S
2. S, U, S
3. U, U, S
4. U, S, S
5. S, U, U

Therefore, we aggregate all paths that lead to a successful third throw:

$P(S, S, S) + P(U, U, S) + P(S, U, S) + P(U, S, S) + P(S, U, U) = 0.448 + 0.072 + 0.144 + 0.072 + 0.144 = 0.748$

To derive this, we consider:

$p_n = P(S,S) + P(U,S)$

Substituting in yields:

By setting $p_n^*$ to equal $p$, we can replace this into the earlier equation:

$p = 0.6 + 0.2p$

This simplifies to

Given that $a_n = p - p_n$, the ratio is computed as follows:

$\frac{a_{n+1}}{a_n} = \frac{p - p_{n+1}}{p - p_n} = \frac{0.75 - p_{n+1}}{0.75 - p_n} = \frac{0.15 - 0.2p_{n}}{0.15} = \frac{1}{5}$

We know that:

Based on the previous calculations, the estimated probability is $p$, which evaluates to 0.75.

Michael's throws are not independent events. The outcome of each throw affects the probability of succeeding on the subsequent throw.